A simple structural estimator of disclosure costs

Abstract

This study recovers a simple firm-level measure of disclosure costs implied by the voluntary disclosure theory of Verrecchia (Journal of Accounting and Economics 12(4), 365–380, 1990). The measure does not require knowledge by the researcher of the distribution of private information and can be implemented with three simple observable inputs: the minimum, average, and frequency of disclosure. We document a positive association of disclosure costs with proxies for existing and potential competition, information asymmetry, and insider trading. Higher values of disclosure costs are associated with lower contemporaneous and future disclosures as well as lower propensity to disclose in holdout samples. Overall, we provide future researchers with an easy-to-implement procedure to structurally estimate unobserved firm-level disclosure costs.

Appendices

Appendix: A

1.1 Asymptotic properties of the estimators

We derive the asymptotic properties of the BBT-cost and NP-cost estimators.

Proposition 1

If c > 0, the estimator c_BBT is consistent ($plim \hat {c}_{BBT}=c$) with asymptotic variance given by

$$ \sqrt{N}(\hat{c}_{BBT}-c)\rightarrow_{d} N(0,\sigma_{BBT}^{2}), $$

where $\sigma _{BBT}^{2}=(H^{\prime }(p))^{2}p (1-p)$ and $H^{\prime }(p)=-\frac {1}{\phi ({\Phi }^{-1}(1-p))}-\frac {\phi ^{\prime }({\Phi }^{-1}(1-p))}{(1-p)(\phi ({\Phi }^{-1}(1-p)))}+\frac {\phi ({\Phi }^{-1}(1-p))}{(1-p)^{2}}$.

Likewise, c_NP is consistent (satisfies $plim \hat {c}_{NP}=c$) with asymptotic variance given by

$$ \sqrt{N}(\hat{c}_{NP}-c)\rightarrow_{d} N(0,\sigma_{NP}^{2}), $$

where $\sigma _{NP}^{2}= \frac {p (p m + m (1 - p))^{2} + (1 - p) p v_{x}} {(1 - p)^{3}}$ and $v_{x}=Var(\tilde {x}|\tilde {x}\geq \tau )$.

The asymptotic variances of the estimators $\hat {c}_{BBT}$ and $\hat {c}_{NP}$ can be easily estimated using sample moments, that is, replacing all elements of the asymptotic variance by their sample estimates, i.e.,

$$ \begin{array}{@{}rcl@{}} \hat{\sigma}_{BBT}^{2}&=&(H^{\prime}(\hat{p}))^{2}\hat{p} (1-\hat{p}),\\ {and} \hat{\sigma}_{NP}^{2}&=& \frac {\hat{p} (\hat{p} \hat{m} + \hat{m} (1 - \hat{p}))^{2} + (1 - \hat{p}) \hat{p} \hat{v}_{x}} {(1 - \hat{p})^{3}}, \end{array} $$

where $\hat {v}_{x}$ is the sample variance of forecasts, then $\hat {\sigma }^{2}_{BBT}$ and $\hat {\sigma }^{2}_{NP}$ are respectively consistent estimators of $\sigma ^{2}_{BBT}$ and $\sigma _{NP}^{2}$.

Proof of Proposition 1:

The proof of consistency is immediate. By continuity, the estimator $\hat {c}_{BBT}$ is consistent. Likewise, by continuity, the estimator $\hat {c}_{NP}$ is consistent, satisfying

$$ plim \hat{c}_{NP}=plim\hat{\tau}+\frac{plim\hat{p}}{1-plim\hat{p}}plim\hat{m}=\tau+\frac{p_{}}{1-p}\mathbb{E}(x|x\geq \tau)=c^{}. $$

Let us derive the asymptotic variances next.

$$ \begin{array}{@{}rcl@{}} \sqrt{N}(\hat{p}-p)\rightarrow_{d} N(0,p(1-p)) \end{array} $$

Applying the Delta method,

$$ \sqrt{N}(\hat{c}_{BBT}-c)\rightarrow_{d} N(0,(H^{\prime}(p))^{2}p (1-p) ), $$

where $H(p)={\Phi }^{-1}(1-p)+\frac {\phi ({\Phi }^{-1}(1-p))}{1-p}$. Taking the derivative of H(p) completes the proof.

We denote x_i as the manager’s information and d_i as the disclosure for each observation i, where d_i = 1 if a forecast is issued or d_i = 0 otherwise.

$$ \hat{c}_{NP}=\hat{\tau}+\frac{\hat{p}}{1-\hat{p}}\hat{m}=\hat{\tau}+\frac{\sum d_{i}/N}{1-\hat{p}}\frac{\sum d_{i} x_{i}}{\sum d_{i}}=\hat{\tau}+\frac{1}{1-\hat{p}}\underbrace{\frac{\sum d_{i} x_{i}}{N}}_{\hat{w}}. $$

In what follows, let us denote $\tilde {x}$ as the random variable, corresponding to the manager’s private information, and $\tilde {d}=1$ if a forecast is issued or d = 0 otherwise. The associated moments to these random variables are denoted $m= \mathbb {E}(\tilde {x}|\tilde {x}\geq \tau )$, $v_{x}= Var(\tilde {x}|\tilde {x}\geq \tau ),$ and $\mathbb {E}(\tilde {d}\tilde {x})=p \mathbb {E}(\tilde {x})=pm$. Note that $\hat {p}$ and $\hat {w}$ are sample means. Therefore, by the central limit theorem,

$$ \sqrt{N}\left( \left( \begin{array}{l} \hat{p} \\ \hat{w} \end{array}\right)-\left( \begin{array}{l} p \\pm \end{array}\right)\right) \rightarrow_{d} N(\mathbf{0}_{2},\underbrace{\left( \begin{array}{ll} Var(\tilde{d}) & cov(\tilde{d},\tilde{d}\tilde{x}) \\ cov(\tilde{d},\tilde{d}\tilde{x}) & Var(\tilde{d}\tilde{x}) \end{array}\right)}_{{\mathbf{V}}_{\mathbf{0}}}. $$

Simplifying this variance-covariance matrix and denoting $m= \mathbb {E}(\tilde {x}|\tilde {x}\geq \tau )$ and $v_{x}= Var(\tilde {x}|\tilde {x}\geq \tau )$,

$$ {\mathbf{V}_{\mathbf{0}}}=\left( \begin{array}{ll} p(1-p) & (1-p)p m \\ (1-p)p m & p(v_{x }+(1-p)m^{2}) \end{array}\right) $$

$$ \begin{array}{@{}rcl@{}} \text{because} Var(\tilde{d})&=& p(1-p);\\ cov(\tilde{d},\tilde{d}\tilde{x})&=& \mathbb{E}(\tilde{d}^{2}\tilde{x})- \mathbb{E}(\tilde{d})\mathbb{E}(\tilde{d}\tilde{x})\\ &=& \mathbb{E}(\tilde{d}\tilde{x})- \mathbb{E}(\tilde{d})\mathbb{E}(\tilde{d}\tilde{x})=(1-p)\mathbb{E}(\tilde{d}\tilde{x})\\ &=& (1-p)p \mathbb{E}(\tilde{x}|\tilde{x}\geq \tau)=(1-p)p m; \end{array} $$

$$ \begin{array}{@{}rcl@{}} \text{and} Var(\tilde{d}\tilde{x})&=& \mathbb{E}(\tilde{d}^{2}(\tilde{x})^{2})- \mathbb{E}(\tilde{d}\tilde{x})^{2}\\ &=& \mathbb{E}(\tilde{d}(\tilde{x})^{2})-(p \mathbb{E}(\tilde{x}|\tilde{x}\geq \tau))^{2}\\ &=& p \mathbb{E}((\tilde{x})^{2}|\tilde{x}\geq \tau)-(p \mathbb{E}(\tilde{x}|\tilde{x}\geq \tau))^{2} \\ &=& p(Var(\tilde{x}|\tilde{x}\geq \tau)+ \mathbb{E}(\tilde{x}|\tilde{x}\geq \tau)^{2})-(p \mathbb{E}(\tilde{x}|\tilde{x}\geq \tau))^{2}\\ &=& p(Var(\tilde{x}|\tilde{x}\geq \tau)+ (\mathbb{E}(\tilde{x}|\tilde{x}\geq \tau))^{2} (1-p))=p(v_{x}+(1-p)m^{2}). \end{array} $$

Next, note that $\hat {c}_{NP}=G(\hat {p},\hat {w})$ where $G(X)=\hat {\tau }+\frac {z}{1-y}$ and X = (y,z). Hence, applying the delta method,

$$ \sqrt{N}(\hat{c}_{NP}^{}-c^{})\rightarrow_{d} N(0,\underbrace{A V_{0} A^{\prime}}_{\sigma_{NP}^{2}}) $$

such that $A=\frac {\partial G}{\partial X^{\prime }}|_{X=(p,pm)}=(\frac {pm}{(1-p)^{2}}, \frac {1}{1-p})$. Therefore,

$$ \begin{array}{@{}rcl@{}} \sigma_{NP}^{2}&=&(\frac{pm}{(1-p)^{2}}, \frac{1}{1-p})\left( \begin{array}{ll} p(1-p) & (1-p)p m \\ (1-p)p m & p(v_{x}+(1-p)m^{2}) \end{array}\right) \left( \begin{array}{l} \frac{pm}{(1-p)^{2}} \\ \frac{1}{1-p} \end{array}\right)\\ &=& \frac {p (p m + m (1 - p))^{2} + (1 - p) p v_{x}} {(1 - p)^{3}}. \end{array} $$

To complete the proof, we further show that $\hat {\tau }$ converges at a rate greater than $\sqrt {N}$. We define J(.), the distribution of x|x > τ and let t > 0, b ∈ (0,p) and α ∈ (1/2, 1). Letting m be the number of disclosures, we can decompose the probability $Prob(N^{\alpha }(\hat {\tau }-\tau )\leq t)$ as follows:

$$ \begin{array}{@{}rcl@{}} \forall t>0,Prob(N^{\alpha}(\hat{\tau}-\tau)\leq t)=Prob(m< bN)Prob(N^{\alpha}(\hat{\tau}-\tau)\leq t|m<bN)\\ +Prob(m> b N)Prob(N^{\alpha}(\hat{\tau}-\tau)\leq t|m>bN+1),\\ \geq Prob(N^{\alpha}(\hat{\tau}-\tau)\leq t|m>[bN])+o(1/N), \end{array} $$

where the inequality follows from the fact that Prob(m < bN) converges to zero as N becomes large and $Prob(N^{\alpha }(\hat {\tau }-\tau )\leq t|m)$ is decreasing in m. The probability in the right-hand side is, up to a negligible term, equal to

$$ \begin{array}{@{}rcl@{}} 1-\left( 1-J\left( \frac{t}{N^{\alpha}}+\tau\right)\right)^{bN} &=&1-\exp \left( bN log\left( 1-J\left( \frac{t}{N^{\alpha}}+\tau\right)\right)\right). \end{array} $$

Taking the limit in $N\rightarrow +\infty ,$

$$ \begin{array}{@{}rcl@{}} lim_{N\rightarrow+\infty}\frac{ log(1-J(\frac{t}{N^{\alpha}}+\tau))}{\frac{t}{N^{\alpha}}}=-\frac{J^{\prime}(\tau)}{1-J(\tau)}=-J^{\prime}(\tau)<0. \end{array} $$

Hence $lim_{N\rightarrow +\infty }1-\exp (bN log(1-J(\frac {t}{N^{\alpha }}+\tau )))= {lim_{N}\rightarrow +\infty }1-\exp (bNt \frac { log(1-J(\frac {t}{N^{\alpha }}+\tau ))}{\frac {t}{N^{\alpha }}})=1$. We conclude: $lim_{N\rightarrow +\infty }Prob(N^{\alpha }(\hat {\tau }-\tau )\leq t)=1$. □

Appendix: B

We prove next that the NP estimator derived under pure disclosure costs continues to hold, even if there is an uncertain information endowment coupled with an inability to credibly communicate a lack of information, as in Dye (1985), or a probability that managers do not care about market perceptions, which would induce them to never engage in costly disclosure.

In addition to a disclosure cost c, the manager may not be informed with some probability q > 0 and then cannot disclose. Although this theoretical assumption is not parsimonious, the reader may note that we are trying to take frictions that may simultaneously exist in the data (Einhorn and Ziv 2008) and hence speak to the fair concern that what we identify as a disclosure cost may, in fact, be uncertain information endowment. In this joint model, the equation for the disclosure threshold remains

$$ \tau-c=\mathbb{E}(x|ND), $$

where the right-hand side is the expected value conditional on not disclosing and is now a function of p as in Jung and Kwon (1988).²⁷

Even though the non-disclosure expectation will be different in this model, we can denote p = (1 − q)(1 − F(τ)), i.e., the probability to disclose, where F(.) is the cumulative probability density of x, and apply the law of total expectations as in the baseline:

$$ \begin{array}{@{}rcl@{}} 0=\mathbb{E}(x)=p\mathbb{E}(x|x\geq\tau)+(1-p)\mathbb{E}(x|ND). \end{array} $$

Solving for E(x|ND) and plugging this equation into (B) implies the same estimator as NP,

$$ c=\tau+\frac{p}{1-p}\mathbb{E}(x|x\geq \tau)). $$

Hence the NP estimator derived under pure disclosure costs continues to hold even if there is uncertainty about information endowment although (as we have shown) this other friction affects the disclosure threshold. The key to this finding is that the probability of disclosure p is endogenous and changes as q increases; we only need to observe this endogenous variable to recover the cost c.

Appendix C: Variable definitions

Table 9